Heat is flowing through a conductor of length | vedclass

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(C) The rate of heat flow through a conductor is given by Fourier's law of heat conduction:$\frac{dQ}{dt} = -KA \frac{dT}{dx}$Since the heat flow rate

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(C) The rate of heat flow through a conductor is given by Fourier's law of heat conduction:$\frac{dQ}{dt} = -KA \frac{dT}{dx}$Since the heat flow rate $\frac{dQ}{dt}$,thermal conductivity $K$,and cross-sectional area $A$ are constants for a uniform conductor,we have:$\frac{dT}{dx} = -\frac{1}{KA} \frac{dQ}{dt} = 	ext{constant}$This implies that the temperature gradient $\frac{dT}{dx}$ is constant.Integrating this expression,we get $T(x) = -mx + c$,where $m$ is a positive constant.This represents a linear decrease in temperature with respect to distance $x$. Therefore,the graph of $T$ versus $x$ is a straight line with a negative slope,which corresponds to Graph $C$.

Heat is flowing through a conductor of length $l$ from $x = 0$ to $x = l$. If its thermal resistance per unit length is uniform,which of the following graphs correctly represents the variation of temperature $T$ with distance $x$?

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